Power maps are abelian group homomorphisms

Let A be an abelian group and fix some k \in \mathbb{Z}. Prove that the map \varphi : A \rightarrow A given by a \mapsto a^k is a homomorphism. If k = -1 show that \varphi is an isomorphism (i.e. an automorphism of A).

\varphi is a homomorphism since \varphi(ab) = (ab)^k = a^k b^k = \varphi(a) \varphi(b) by a previous exercise.

Now fix k = -1. \varphi is surjective since for all a \in A, a = \varphi(a^{-1}). Suppose now that \varphi(a) = \varphi(b); then a^{-1} = b^{-1}, so that a = b and \varphi is injective. Thus with k = -1, \varphi is an automorphism.

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